What are the solutions to the equation 4((5-3y)/2)^2-3((5-3y)/2)y+9y^2=15 ?

The solutions to the equation 4((5-3y)/2)^2-3((5-3y)/2)y+9y^2=15 are y= 4/3 ,y= 1/3

Find the zeros of 4((5-3y)/2)^2-3((5-3y)/2)y+9y^2=15

The zeros of 4((5-3y)/2)^2-3((5-3y)/2)y+9y^2=15 are y= 4/3 ,y= 1/3

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4 (\frac{5 - 3 y}{2})^{2} - 3 (\frac{5 - 3 y}{2}) y + 9 y^{2} = 15

y = \frac{4}{3}, y = \frac{1}{3}

Decimal Notation

y = 1.33333 \dots, y = 0.33333 \dots

Solution steps

4 (\frac{5 - 3 y}{2})^{2} - 3 (\frac{5 - 3 y}{2}) y + 9 y^{2} = 15

Expand

4 (\frac{5 - 3 y}{2})^{2} - 3 (\frac{5 - 3 y}{2}) y + 9 y^{2} : \frac{45 y ^{2}}{2} - \frac{75 y}{2} + 25

\frac{45 y ^{2}}{2} - \frac{75 y}{2} + 25 = 15

Move

15

to the left side

\frac{45 y ^{2} - 75 y}{2} + 10 = 0

Write in the standard form

a x^{2} + b x + c = 0

\frac{45 y ^{2}}{2} - \frac{75 y}{2} + 10 = 0

Solve with the quadratic formula

y_{1, 2} = \frac{- ( - \frac{75}{2} ) \pm ( - \frac{75}{2} ) ^{2} - 4 \cdot \frac{45}{2} \cdot 10}{2 \cdot \frac{45}{2}}

(- \frac{75}{2})^{2} - 4 \cdot \frac{45}{2} \cdot 10 = \frac{45}{2}

y_{1, 2} = \frac{- ( - \frac{75}{2} ) \pm \frac{45}{2}}{2 \cdot \frac{45}{2}}

Separate the solutions

y_{1} = \frac{- ( - \frac{75}{2} ) + \frac{45}{2}}{2 \cdot \frac{45}{2}}, y_{2} = \frac{- ( - \frac{75}{2} ) - \frac{45}{2}}{2 \cdot \frac{45}{2}}

y = \frac{- ( - \frac{75}{2} ) + \frac{45}{2}}{2 \frac{45}{2}} : \frac{4}{3}

y = \frac{- ( - \frac{75}{2} ) - \frac{45}{2}}{2 \frac{45}{2}} : \frac{1}{3}

The solutions to the quadratic equation are:

y = \frac{4}{3}, y = \frac{1}{3}

x^{2} - 4.4 x = - 3.84

z^{2} + 8 = 0

4.905 x^{2} - 12 x - 70 = 0

v, K E = \frac{1}{2} m v^{2}

x^{2} = 16 x - 64

What are the solutions to the equation 4((5-3y)/2)^2-3((5-3y)/2)y+9y^2=15 ?
The solutions to the equation 4((5-3y)/2)^2-3((5-3y)/2)y+9y^2=15 are y= 4/3 ,y= 1/3
Find the zeros of 4((5-3y)/2)^2-3((5-3y)/2)y+9y^2=15
The zeros of 4((5-3y)/2)^2-3((5-3y)/2)y+9y^2=15 are y= 4/3 ,y= 1/3